P1 and P3 are fixed points. P2 lies on the line perpendicular to P1P3 through P3. The sequence P4, P5, P6, ... is defined inductively as follows: Pn+1 is the foot of the perpendicular from Pn to Pn-1Pn-2. Show that the sequence converges to a point P (whose position depends on P2). What is the locus of P as P2 varies?
Solution
PnPn+1Pn+2 lies inside Pn-1PnPn+1. So we have sequence of nested triangles whose size shrinks to zero. Each triangle is a closed set, so there is just one point P in the intersection of all the triangles and it is clear that the sequence Pn converges to it.
Obviously all the triangles PnPn+1Pn+2 are similar (but not necessarily with the vertices in that order). So P must lie in the same position relative to each triangle and we must be able to obtain one triangle from another by rotation and expansion about P. In particular, P5P4P6 is similar (with the vertices in that order) to P1P2P3, and P4P5 is parallel to P1P2, so the rotation to get one from the other must be through π and P must lie on P1P5. Similarly P3P4P5 must be obtained from P1P2P3 by rotation about P through π/2 and expansion. But this takes P1P5 into a perpendicular line through P3. Hence P1P is perpendicular to P3P. Hence P lies on the circle diameter P1P3.
However, not all points on this circle are points of the locus. P3P5 = P3P4 cos P1 = P3P1 sin P1 cos P2 = 1/2 P3P1 sin 2P1, so we can obtain all values of P3P5 up to P1P3/2. [Of course, P2, and hence P5, can be on either side of P3.]. Thus the locus is an arc XP3Y of the circle with XP3 = YP3 and ∠XP1Y = 2 tan-11/2. If O is the midpoint of P1P3, then O is the center of the circle and ∠XOY = 4 tan-11/2 (about 106o).
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002